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15th - 1 1.7 P)VVICR SERIES 1 11β€”1 2 lnn Uβ€œβ€”β€œ< 1...

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Unformatted text preview: 1 1.7 P( )VVICR SERIES . _ 1; 11β€”1 _ 2. lnn Uβ€œβ€”β€œ < 1 2:Β» lnn (x l < 1 5β€˜ < 1 6 < X < 4', when X = β€”(3 we have 11 β€”1 oc- Us 11 β€”1 oc- (X1 .1β€œ TX! 7:: z (β€”1)β€œ, a divergent series; when X = β€”4 we have 2: 1, a divergent series not 1171 (a) the radius is 1; the interval of convergence is β€”6 < X < β€”4 (b) the interval of absolute convergence is 76 < X < 74 (c) there are no values for which the series converges conditionally - 11.,+ - (xβ€”2]"_l 1on xβ€”2 5.1111111” uβ€”"J<1=>nlll]}x__l W'W 2:Β» β€”8 < X < 12; when X = β€”8 we have (β€”1)β€œ, a divergent series; when X = 12 we have 1, a divergent 11β€”1 11β€”] series (a) the radius is 10; the interval of convergence is β€”3 < X < 12 (b) the interval of absolute convergence is β€”3 < X < 12 (c) there are no values for which the series convelges conditionally 1) - uniβ€œ - 10'β€œ 7 ntlnnli - r1 - lnn β€˜ 27' llllpbc- 1.1.. <1 2:} 111β€”19139 (n11](lntn1]1_β€˜1"β€˜ K" < l 2:} lxl (nlβ€”lpbc- n11) lntntlJ) < l 1. 3 2 =>|X|(1) lim_ (1’) <1=>|X|(li1n_m1) <1=>|X|<1=>β€”1<X<l;whenX=β€”lwehave [lβ€”β€˜CXJ 11β€”β€˜09 n TI: TI: (β€”1 1" - _. ,. _ _. 1 - . E l wlnch conv erges absolutel) , when X β€” 1 we ha\ e E m Wthll converges n_ nβ€”l (a) the radius is 1; the interval of convergence is 71 < X < 1 (b) the interval of absolute convergence is β€”1 < X < 1 (c) there are no values for which the series convelges conditionally 2n+3 xβ€” n 32. Hut β€œ1—” <1 2:Β» li1n 145] 11β€”100 Us 11β€”100 ( I) 2 _ 2 . ::>(x.β€”:/β€”]<1::>(Xβ€”\,/Β§) <2::> Xβ€”\/Β§<\/Β§:Β’>β€” 6<Xβ€”\/Β§<\/Β§:Β’>0<X<2\/Β§;when 211β€” x’ β€˜1Β» 2 3Β° :1 1 2 3Β° _ , , . . X = Owe have 2: $ = β€” Z 2 gβ€œ = β€” Z we which diverges s1nce β€œ11111)C an $ 0; when X = EVE nβ€”1 nβ€”1 nβ€”l β€˜ I 2n+l so ,5 so so \, . n 12 . . we have 2: ( 2]" = Z 3 gβ€œ = 2: V6, a d1vergent series nβ€”l nβ€”l 11β€”] (a) the radius is V5; the interval of convergence is 0 < .X < 2V5 (b) the interval of absolute convergence is 0 < X < 2V5 (c) there are no values for which the series convelges conditionally . ml 2 s - ., 37.nliβ€˜mxl β€œ:1: <1 => β€œ11ng (β€œ13”) 1%)β€œ <1=>("3”-’n1_11nw|1|<1=> *3” <1 => x~<2 2:Β» < x/E 2:Β» β€” G < .X < x/E; at X = ; VG we have (1)rl which diverges; the interval of convergence is nβ€”Cl x. _l n β€” 6 < X < x/E ; the series 2: (k 3' 1) is a convelgent geometric series when β€”\,/E < .X < V5 and its sum is 1-1β€”0 11.8 TAYLOR AND MACIAURIN SERIES 3. fix) 2 3? = f], f’(x) = β€”x_2, f"(x) 2 211β€”3, f’"(:x) = β€”6x_4; f(2) , f’(2) = β€” ET, f”(2) = 1:, f’”(x) = β€” =>P.,(x)=%,P](x)=%β€”}T(xβ€”2),P3(x)=%β€”3T(xβ€”2)β€”%(xβ€”2)', P;1(x)=?3 11x 2) 11x 213 3,11: 21" 6. f(x)=cosx,f’(x)=β€”sinx,f"(x)=β€”cosx,f’”(x)=sinx;f(Β§]=eosgzfi, PG] :β€”sin%:β€”7]E,f”(%] :β€”e05Β§:β€”7]E,f’”(%] :sinfizfi 2:Β» P1100: V12, P100=fig-figb-fiththfi-figfi-fl-ޤ(x-%Ji 2 :1 flatwareβ€”452%) β€”,3β€”51xβ€”11 12. fix): (1 β€”x)_] => fix): (1 β€”x)β€˜3.f”(x) = 211β€” 111β€”3, 1%) = 3:11 β€” 111β€”4 => 1mm 2 ma β€” 11-11-2110) 2 1, 1'10) 2 1, 9’10) 2 2, 1111(0) = 3:, ,fmi-(m = k! '35 =1β€”xβ€”x2β€”Xβ€œβ€”... =2: xβ€œ :8 1β€”K n-D fix) 2 3xβ€œ β€” x4 β€” 2.313 β€” x2 β€” 2 2:Β» f’(x) = 15.31β€œ1 β€” 4x3 β€” 6x2 β€” 2x, f"(x) = 60113 β€” 12.313 β€” 12xβ€” 2, 1mm 2 180112 β€” 24.x β€” 12,1001β€œ: 3601: β€” 24, Β£15101) 2 360, Mm = 0 ifn :-~ 6; f(β€” 1) = β€”7, 91β€”11 = 23,1"1β€”1): β€”82,f”’(β€”1) = 216, 1(411β€”1) = β€”334, 1(5?(β€” 1) = 360, 1031β€”1) 2 Oifn :5 6 24. =1Β» 3.1β€œ x4 2113 x3 2 = 7 23(11 1) 3%: 1)2 231,601 11" 343,411: 1)4 355,001 1)β€œ = 7 231’): 1) 41(11 1].2 36(x 1)?β€˜ 1601 1)β€œ1 3(11 1)β€œ 11.9 CONVERGENCE OE TAYLOR SERIES; ERROR liS'lβ€˜IPB/IA'IES . L2 2" _ _ 1:211 ; 113 35 1β€”D" _ n31: 6- = 1111;: j (5β€”,) = ((1%) ) = 2;] 1 = z 1']β€” Flβ€” nβ€” : 1 _ _ x_β€œ _ x_"' _ 2-2I 21-4! 24-6! 15 If; : Kenβ€”121]: K3 Z;an : Zoznxnm : X3 _ 2x3. _23K4 _ 23.x". _ 22 V l β€” x = 1 β€” 5, β€” g: β€” β€” . By the Alternating Series Estimation Theorem the |err01'| < % < @ = 1.25 X 10β€”β€œ 33. [5111'] 1: when x = %;the sum is tanβ€”1 :3 0.808448 11.10 APPLICA'Iβ€˜IONS OE POWER SERIES ._--) . _1 3 1’ _L _2 _Β§':1" , 7- [l_xβ€˜..,]1;-=l Fifi ( .,](2!]1x) (Β£11 1;!( 111x) :1 13x1 gxh 156x: 13. (1 β€” 211)3 = 1 31 2x) (MELβ€”2""! 13β€œβ€œ131'2” = 1 6x 12112 3114* 24. [1β€” x3] 3/ β€” 211)? = a] (2513 2am): (3613 251] 511))? (4:14 2:13 233))(3 (narl nan_g)xβ€œ = 0 2:Β» a] = 0, 2:13 β€” 2aβ€œ 2 0, 333 β€” 351] = 0, 4:14 β€” 4513 = 0 and in general naJ1 β€” nan_2 = 0. Since y = 3 when x: 0,we]1a\-'e aβ€œ = 3. Therefore :13 = β€”3,ay, 20,34 = 3, ,Elgm] 20,512rl = (β€”1)β€œ3 ,_ 2 .1 _ 3Β° 2 _ 3Β° 2 n_ 3 2:Β» }β€”3β€”3x β€”3x β€”... β€”Z%)3(β€”l)β€œxβ€œβ€”Z%3[β€”x] β€”m_r A 3 _1_ 31. }”—x3y=233 633): (4-33.; am}? (n(n Dan an_4)xβ€œβ€˜2β€” . 2x 2:Β» 233:0,63321, 4 - 334 β€” an 2 0, 5 - 435 β€” a] = 0, and in general n(n β€” Dan β€” an_4 = 0. Since 3/ = b and y = a when x = CI, wehavean : and a] = b. Therefore :12 = 0,02. = = β€” fins : β€” 4%β€œ -. 20,617 = ifβ€œ , _ 1 :1. a 4 b r. 1 7 2m" bx" 2* .β€˜ β€” a b" 2-3 X 3-4 x 4-5 X 2-3-07 9β€˜ 3 4 s 4 5-3-9 F(X):J:([2_%_%_%_HI)dt:[In 1|] 1L:- ;. _ :Β» |e1'1'01'| < :5 0.000013 54.0: l)sin(x]}]]=(x 1)(X1H 1 1 .β€˜,)=1_ 1 20m 11:1 S!(x} 11-1 you 1:? _ sum 11' _ ::>xlip})cl(xβ€”l)5in(x:]]: lim_l (1 1 1 .mjzl ...
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15th - 1 1.7 P)VVICR SERIES 1 11β€”1 2 lnn Uβ€œβ€”β€œ< 1...

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